It is well known that the digits of multiples of nine sum to nine; i.e.,
9→9, 18→1+8=9, 27→2+7=9, . . ., 99→9+9=18→1+8=9, 108→1+0+8=9,
etc. Less well known is that the sum of digits of multiples of other numbers have simple
patterns although not so simple as the case of nine. These are shown below:

Number

Repeating Cycleof Sum of Digitsof Multiples

2

{2,4,6,8,1,3,5,7,9}

3

{3,6,9,3,6,9,3,6,9}

4

{4,8,3,7,2,6,1,5,9}

5

{5,1,6,2,7,3,8,4,9}

6

{6,3,9,6,3,9,6,3,9}

7

{7,5,3,1,8,6,4,2,9}

8

{8,7,6,5,4,3,2,1,9}

9

{9,9,9,9,9,9,9,9,9}

10

{1,2,3,4,5,6,7,8,9}

11

{2,4,6,8,1,3,5,7,9}

12

{3,6,9,3,6,9,3,6,9}

13

{4,8,3,7,2,6,1,5,9}

It is asserted that the sum of digits follows a repeating sequence of length 9. This is
so because for any decimal representation x, the number 10*x (ten times x) will have the same sum of digits.
Multiplication by the base number 10 simply moves the
digits one place to the left and puts a zero in the units place. More generally,

DigitSum(10k*n) = DigitSum(n).

This operation, multiplication by the k-th power of 10, moves the digits k places to the
left and places k zeroes on the right. The sum of digits is the same.

Less obviously adding a nine at any place in a decimal representation reduces the digit by one and adds one to the
digit in the next higher place, and thus the sum of the digits is not altered. That is to say,

DigitSum(n+9*10k) = DigitSum(n).

Going back to the above table, what is immediately suggested by the information presented there is that
the sequence for a multiple digit number m is the sequence for the sum of the digits of m,
the digit sum of m.
For example, the sequence for 12 is the same as the sequence for 1+2=3. Likewise the
sequence for 13 is the same as the sequence for 1+3=4. This of course also holds true for
10.

The sequences for 3 and 6 are composed of the subsequences {3,6,9} and {6,3,9}, respectively.
Thus the sequences for these digits have three copies of the subsequences of length 3. The
sequence for 9 is nine copies of a subsequence of length 1. Note that {3,6,9}=3*{1,2,3} and that
{6,3,9}=3*{2,1,3}. It is also worth noting that for all digits except 3, 6 and 9 the length of the sequence is
nine and that for 3, 6 and 9 the lengths of their sequences, 3 and 1, are factors of 9.

The structures of the length 9 sequences are noteworthy. Here is a naive perception of the structure
of the sequences. For 2 the sequence is the even
digits in ascending order then the odd digits also in ascending order. For 4 the sequence
is two generally descending sequences interleaved; i.e., 4,3,2,1,9 with 8,7,6,5 interleaved.
For 5 the sequences is also two interleaved sequences but ascending rather than descending; i.e.,
5,6,7,8,9 with 1,2,3,4 interleaved. For 7 the sequence is a descending sequence of the odd
digits starting with 7 then the even digits in descending order with 9 the odd digit left
out coming last. For 8 the sequence is a single descending sequence of digits starting
with 8 and finishing with the digit 9 which was left out of the descending sequence.
Later a more mathematical explanation of the structure of the sequences will be given.

The DigitSum Function

Let n be a number and let Dec10(n) be the decimal representation of n.
Let DigitSum(z) be the ultimate sum of digits of a decimal representation z; i.e., if the
sum of digits of a decimal representation is greater than nine then the sum of that number's
digits
is computed until a single digit is ultimately obtained.

The proposition that DigitSum(x*y)=DigitSum(DigitSum(x)*y)) establishes that the sequences for the
multiples of 12 and of 13 are the same as the sequences for 3 (1+2) and 4 (1+3),
respectively.

Decimal Representation of Numbers

In order for the following to make sense one must stop thinking of a number in terms of
its decimal representation and think of a number in terms of an appropriate number of tally
marks so three would be (|||) and eleven (||||||||||||).

Consider how one obtains the decimal representation of a number. To get the last digit one
divides the number by ten and takes the remainder as the last digit. That last digit is
subtracted from the number and the result divided by ten. Then the decimal representation of
that quotient is sought. The process is repeated and the next the last digit is obtained.

An alternate characterization of the process of finding the decimal representation of a number is
that the k-th power digit for a whole number n is:

ck = (trunc[n/10k])%10

where trunc[] means the fractional part is thrown away and m%10 means the remainder after
division by 10. In terms of the terminology from the programming language Pascal the
formula is

ck(n) = (n div 10k) mod 10

The sum of the digits for a number n is then

Sum = Σk ck(n),

but this is not necessarily the digit sum for the number. The process has to be repeated
iteratively on the sum of the digits.

It was previously noted that for any two digits whose sum is greater than ten,

DigitSum(a+b) = 1 + (a+b−10) = a+b−(10−1) = a+b−9

In general then for any decimal representation x

DigitSum(x) = Sumofdigits(x) − m*9
where m is such that DigitSum(x) is reduced to a single digit.

Another way of expressing this is that the DigitSum for a number n is simply the
remainder after division by 9; i.e., DigitSum(n)=(n%9). DigitSum arithmetic is simply
arithmetic modulo 9.

For comparison the multiplication table for modulo 9 arithmetic is:

Multiplication Table for Modulo 9 Arithmetic

0

0

0

0

0

0

0

0

0

<

0

1

2

3

4

5

6

7

8

0

2

4

6

8

1

3

5

7

0

3

6

0

3

6

0

3

6

0

4

8

3

7

2

6

1

5

0

5

1

6

2

7

3

8

4

0

6

3

0

6

3

0

6

3

0

7

5

3

1

2

6

4

2

0

8

7

6

5

4

3

2

1

If the 0's were replaced by 9's and the table rearranged so the first column becomes the last column and the first row
becomes the last row the result would be
identical to the table for the sequences of digit sums.

The Rearrangement of the
Multiplication Table for Modulo 9 ArithmeticWith 9 Substituted for 0

1

2

3

4

5

6

7

8

9

2

4

6

8

1

3

5

7

9

3

6

9

3

6

9

3

6

9

4

8

3

7

2

6

1

5

9

5

1

6

2

7

3

8

4

9

6

3

9

6

3

9

6

3

9

7

5

3

1

2

6

4

2

9

8

7

6

5

4

3

2

1

9

9

9

9

9

9

9

9

9

9

Some Proofs

The proof of the property that the DigitSum of any multiple of nine is equal to nine starts with the obvious proposition

DigitSum(10*n) = DigitSum(n).

From this it follows from the proposition that
DigitSum(x−y)=DigitSum(x)−DigitSum(y) that

DigitSum((10−1)*n) = 0
and thus
DigitSum(9*n) = 0
but in modulo 9 arithmetic
0 is the same as 9, so
DigitSum(9*n) = 9

The way to understand the sequence for 8 is that when 8 is added to any digit except 0 or 1
in any place the digit is reduced by 2 and one added to the digit of the next place. This
results in a net decrease in the sum of digits of one. Thus when 8 is added to 8 the digit
becomes 6, a reduction of 2, and 1 is added to the next higher digit, a net decrease in the
sum of the digits of 1. Thus added 8 to 8 results in a sum of digits of 7. Adding 8 to 7
results in a sum of digits of 6, and so on down to adding 8 to 1 which gives 9. Even this
fits into the rule in the sense that if 1 were reduce by 1 the result would be 0 which is
equivalent to 9 modulo 9. Likewise adding 7 to a digit reduces it by 3 and adds 1 to the
digit in the next place, a net reduction in the sum of digits of 2. Thus when 7 is added to 7
the sum of digits is reduced to 5. When 7 is added to 5 the sum of digits is reduced to 3.
When 7 is added to 3 the sum of digits is reduced to 1. If 7 is added to 1 the result is 8,
but that 8 could be consider as a reduction of 1 by 2 modulo 9. The addition of 7 to 8 results
in a sum of digits of 6 and so on down to a sum of digits of 2. The addition of 7 to 2
results in a sum of digits of 9, but that 9 can be considered 0 modulo 9 and thus it is
a reduction of 2.

Generalization to Other Number Bases

There is nothing special about 9; it is simply the number base ten less one. The digitsum sequences for multiples in the
hexadecimal (base 16) number system is:

Number

Repeating Cycleof Sum of Digitsof Multiples

2

{2,4,6,8,a,c,e,1,3,5,7,9,b,d,f}

3

{3,6,9,c,f,3,6,9,c,f,3,6,9,c,f}

4

{4,8,c,1,5,9,d,2,6,a,e,3,7,b,f}

5

{5,a,f,5,a,f,5,a,f,5,a,f,5,a,f}

6

{6,c,3,9,f,6,c,3,9,f,6,c,3,9,f}

7

{7,e,6,d,5,c,4,b,3,a,2,9,1,8,f}

8

{8,1,9,2,a,3,b,4,c,5,d,6,e,7,f}

9

{9,3,c,6,f,9,3,c,6,f,9,3,c,6,f}

a

{a,5,f,a,5,f,a,5,f,a,5,f,a,5,f}

b

{b,7,3,e,a,6,2,d,9,5,1,c,8,4,f}

c

{c,9,6,3,f,c,9,6,3,f,c,9,6,3,f}

d

{d,b,9,7,5,3,1,e,c,a,8,6,4,2,f}

e

{e,d,c,b,a,9,8,7,6,5,4,3,2,1,f}

f

{f,f,f,f,f,f,f,f,f,f,f,f,f,f,f,f}

In general, for numbers of base b the digit sums are equivalent to arithmetic
modulo (b−1).
The sequences which contain subsequences are the multiples of the factors of (b−1)
other than unity.
In the case of b=ten these factors are 3 and 9. Therefore there are subsequences for
{3,6,9}. The lengths of the subsequences are (b−1) divided by the factors; i.e.,
in the case of b=ten 3 and 1.

For b=sixteen the factors are three, five
and fifteen and so subsequences occur for {3,6,9,c,f,5,a} and the lengths of the
subsequences are 3, 5 and 1.